Since the particle is initially at rest, the net force acting on it is zero. This means that the forces $$F_1$$, $$F_2$$, and $$F_3$$ are balanced. Given that $$F_2$$ and $$F_3$$ are acting perpendicularly, we can represent the balance of forces as follows:

$$F_1 = \sqrt{F_2^2 + F_3^2}$$

We can now calculate the equivalent force resulting from $$F_2$$ and $$F_3$$:

$$F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}$$

Since the particle is initially at rest, when the force $$F_1$$ is removed, only $$F_2$$ and $$F_3$$ remain. The net force acting on the particle is the equivalent force resulting from $$F_2$$ and $$F_3$$, which we have calculated to be $$10 \mathrm{~N}$$.

Now, we can use Newton's second law to find the acceleration of the particle:

$$F = ma$$

Where:

• $$F$$ is the net force acting on the particle
• $$m$$ is the mass of the particle
• $$a$$ is the acceleration of the particle

Rearranging the equation to solve for $$a$$:

$$a = \frac{F}{m}$$

Plugging in the values for $$F$$ and $$m$$:

$$a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}$$

The acceleration of the particle when the force $$F_1$$ is removed is $$2 \mathrm{~ms}^{-2}$$, which corresponds to Option C.